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3 tháng 9 2019

 | x+1|=0                                        b) sai đè nha bn             

=> x+1=0                                                                                

=> x=0-1

=>x=(-1)

3 tháng 9 2019

2

b) \(\frac{50}{51}>\frac{50}{58};\frac{50}{58}>\frac{49}{58}\)=> \(\frac{50}{51}>\frac{49}{58}\)

c)  vì \(\frac{2019}{2018}>1\)=> \(\frac{2019+1}{2018+1}=\frac{2020}{2019}< \frac{2019}{2018}\)

B= 1/1.2+1/2.3+...+1/2019.2020

B=1/1-1/2+1/2-1/3+...+1/2019-1/2020

B=1-1/2020=2020/2020-1/2020=2019/2020

14 tháng 5 2019

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\) 

Với  :   \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\) 

Và   :   \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\) 

             \(B=1-\frac{1}{2020}< 1< A\)

22 tháng 5 2019

đặt 22018 = a ; 32019 = b ; 52020 = c

Ta có : \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)

\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(2B=\frac{2}{1.2}+\frac{2}{3.4}+...+\frac{2}{2019.2020}\)

\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)

\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)

\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}=1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)

\(B< \frac{3}{4}\)

\(\Rightarrow A>1>\frac{3}{4}>B\)

22 tháng 5 2019

Mình chỉ biết cách tính B thôi, đây nhé:

B= \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)

B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(B=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2020}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-2\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2019}+\frac{1}{2020}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1010}\right)\)

\(B=\frac{1}{1011}+\frac{1}{1012}+....+\frac{1}{2019}+\frac{1}{2020}\)

AH
Akai Haruma
Giáo viên
14 tháng 5 2019

Lời giải:

\(B=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{2019.2020}\)

\(\Rightarrow 2B=\frac{2}{1.2}+\frac{2}{3.4}+\frac{2}{5.6}+....+\frac{2}{2019.2020}\)

\(< 1+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{2018.2019}+\frac{1}{2019.2020}\)

\(2B< 1+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+....+\frac{2019-2018}{2018.2019}+\frac{2020-2019}{2019.2020}\)

\(2B< 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\( 2B< 1+\frac{1}{2}-\frac{1}{2020}< 1+\frac{1}{2}\)

\(B< \frac{3}{4}\)

---------------------

Đặt \(2^{2018}=a; 3^{2019}=b; 5^{2020}=c(a,b,c>0)\)

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}> \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1\)

\(\Rightarrow A>1> \frac{3}{4}> B\)

15 tháng 5 2019

thầy giải hay quá

26 tháng 5 2020

mình nhầm , thay 2019 = 2020 nhé

3 tháng 1 2018

\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)

\(\Leftrightarrow\left(\frac{x-4}{2021}+1\right)+\left(\frac{x-3}{2020}+1\right)=\left(\frac{x-2}{2019}+1\right)+\left(\frac{x-1}{2018}+1\right)\)

\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}=\frac{x+2017}{2019}+\frac{x+2017}{2018}\)

\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}-\frac{x+2017}{2019}-\frac{x+2017}{2018}=0\)

\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)=0\)

Mà \(\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)\ne0\)

\(\Leftrightarrow x+2017=0\)

\(\Leftrightarrow x=-2017\)

Vậy ..

3 tháng 1 2018

=> (x-4/2021 +1) + (x-3/2020 +1) = (x-2/2019 +1)+ (x-1/2018 +1)

=> x+2017/2021 + x+2017/2020 = x+2017/2019 + x+2017/2018

=> x+2017/2018 + x+2017/2018 - x+2017/2020 - x+2017/2021 = 0

=> (x+2017).(1/2018+1/2019+1/2020+1/2021) = 0

=> x+2017 = 0 ( vì 1/2018+1/2019+1/2020+1/2021 > 0 )

=> x=-2017

Vậy x=-2017

k mk nha

17 tháng 1 2017

Bài 1:Áp dụng C-S dạng engel

\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)

\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)